This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Taught By

Dr. Bonnie H. Ferri

Professor

Dr. Joyelle Harris

Director of the Engineering for Social Innovation Center

Dr Mary Ann Weitnauer

Transcript

The topic of this problem is nodal analysis, and we're going to solve circuits with dependent voltage sources. The problem is to find the current I sub 0 in the circuit shown below. If we look at that circuit, we have on the left-hand side a current controlled voltage source where the current is I sub x and is defined as the current on the right-hand side of the circuit. That's our I sub x. So we have a current controlled voltage source. Ultimately, we're looking for the current I sub 0, which is through the 2 kiloohm resistor at the top of the circuit. So we're going to solve this problem using nodal analysis. That means that we're going to use Kirchhoff's current laws and that the first step in our process is to identify our nodes. So we going to use Kirchhoff's current law, and we're going to solve using the currents flowing out of the nodes. So we're going to sum the currents flowing out of each of those nodes. First of all, identifying the nodes, we have a node 1 at the upper left-hand side of the circuit. We have a node 2 at the upper right-hand side of the circuit. We have a ground node at the bottom of the circuit, which is our reference node at 0 volts. So we're going to sum, again, the currents about each one of those nodes, starting with node 1. Node 1 has the current associated with the dependent source flowing into it. It has the current flowing through the 2 kiloohm resistor from the bottom to the top. And it has the current flowing through the 2 kiloohm resistor at the top of the circuit as well. So we have these three paths for current into node 1. But we know with nodal analysis that what we're really after is we're after the nodal voltages. So, in this problem, if we try to sum the currents into node 1 and we sum the current from the dependent source, we'd be introducing an additional variable. It would be the current I 2K I sub x. So it would be the current through the dependent source. This would introduce an additional unknown variable. So we would have the two unknown nodal voltages, V1 and V2, which we can write equations about. And we'd have the additional, the current associated with the current flowing through the dependent source. So that would introduce a third variable. If we did it that way, we'd find that we have two equations that we could write for the nodal equations, and three unknowns. And we wouldn't have enough equations to solve for all of the unknowns. So, what we notice on this problem is that that dependent source is tied to a reference point or ground at the bottom of the circuit and at the top to the nodal voltage V1. So instead of writing the currents about that node, we can go directly for the nodal voltage, which is ultimately what we're after in the first place. And we would do that by writing the equation V1, because V1 is the positive polarity of the dependent source, minus the ground, which is tied to the negative polarity of the source, is equal to 2K I sub x. That's our first equation. We have a second equation for node 2. And node 2 has three currents flowing out of it. First of all, we know we have -4 milliamps, because the current source is flowing into node 2. We have the current which is associated with the 2 kiloohm resistor at the top flowing out. It's going to be V2 minus V1 divided by 2K. And we have the current which is flowing through the 1 kiloohm resistor on the right-hand side of the circuit into node, or sorry, we're summing out, we have this current flowing out of node 2 through the 1 kiloohm resistor down to the reference node at the bottom. So that's going to be V2 minus 0 divided by 1K. And that's all of our currents, and they sum up to be 0. So we have these two equations and two unknowns where we have V1 and V2 unknown. And if we look closely, we also have the I sub x, which is part of our dependent voltage source. It's the controlling parameter for our dependent voltage source. So that adds our third unknown to our set of two equations. So we need one more equation which relates the controlling parameter I sub x to the nodal voltages. And so we can write that third equation, because we know that I sub x is the current flowing down through the 1 kiloohm resistor on the right-hand side of the circuit. So V2 minus 0 for the ground node, V2 minus 0 divided by 1 kiloohm gives us the current down through that 1 kiloohm resistor, divided by 1K is equal to I sub x. So now we have three equations and we have three unknowns. And since we have those, we can then solve for our unknown I sub 0. I sub 0, as we see, flows between node 1 and node 2. So we need to be able to find V1 and V2 for the nodal voltages to find I0 using Ohm's law. So if we solve this set of equations for V1 and V2, we have a V1 which is equal to 16 volts. We have a V2 which is equal to 8 volts. And we know that we can find I0 from that, because I0 is equal to V1 minus V2 divided by 2K. And so that's 16 minus 8 divided by 2 kiloohms. And so ultimately, I0 will come out to be 4 milliamps.

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